Question

A body has a mass of \(12.6 \mathrm{kg}\) and a speed of \(0.87 c .\) (a) What is the magnitude of its momentum? (b) If a constant force of \(424.6 \mathrm{N}\) acts in the direction opposite to the body's motion, how long must the force act to bring the body to rest?

Short answer

(a) The momentum magnitude is approximately \(22.05c\). (b) Time to stop is about \(15580\) seconds.

Step-by-step solution

Understanding Relativistic Momentum Formula

To find the momentum of an object moving at a significant fraction of the speed of light, we use the relativistic momentum formula: \( p = \gamma m v \), where \( \gamma \) is the Lorentz factor: \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \).

Calculate the Lorentz Factor

Given that the speed \( v = 0.87c \) and \( c \) is the speed of light, calculate the Lorentz factor \( \gamma \):\[\gamma = \frac{1}{\sqrt{1 - (0.87)^2}} \approx \frac{1}{\sqrt{1 - 0.7569}} \approx \frac{1}{0.4982} \approx 2.0079.\]

Calculate Relativistic Momentum

Using \( \gamma \approx 2.0079 \), mass \( m = 12.6 \mathrm{kg} \), and speed \( v = 0.87c \), find the momentum: \[ p = \gamma m v = 2.0079 \times 12.6 \times 0.87c \approx 22.05c. \]

Apply Newton’s Second Law for Time Calculation

To stop the body, use the impulse-momentum theorem \( F \cdot t = -p \) to find the time \( t \) it takes: \[ t = \frac{-p}{F} = \frac{-22.05c}{424.6}. \]

Plug in Values and Solve

Substitute \( c \approx 3 \times 10^8 \mathrm{m/s} \) into the equation for \( t \):\[p = 22.05 \times 3 \times 10^8 \approx 6.615 \times 10^9 \ t = \frac{-6.615 \times 10^9}{424.6} \approx -15580 \mathrm{s}.\]The negative sign indicates a direction opposite to initial momentum.

Key concepts

Lorentz Factor

The Lorentz factor, denoted as \( \gamma \), is a crucial quantity in the realm of relativistic physics. It measures how much time, length, and relativistic momentum differ from their classical mechanics counterparts when an object is traveling at a significant fraction of the speed of light. This factor becomes especially important in scenarios where velocities approach the speed of light, \( c \).

The formula to calculate the Lorentz factor is:

• \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \)

Here, \( v \) is the velocity of the object, and \( c \) is the speed of light in a vacuum, roughly \( 3 \times 10^8 \) meters per second.

The value of \( \gamma \) increases as \( v \) approaches \( c \), meaning objects moving very fast experience more pronounced relativistic effects. When interpreting \( \gamma \), remember:

• As \( v \to c \), \( \gamma \to \infty \).
• If \( v = 0 \), then \( \gamma = 1 \), resembling classical physics scenarios.

The Lorentz factor is directly tied to changes in momentum, time dilation, and length contraction, all crucial for understanding relativistic momentum.

Impulse-Momentum Theorem

The impulse-momentum theorem is a fundamental principle that connects forces to changes in an object's momentum. This theorem is expressed as:

• \( F \cdot t = \Delta p \)

Where \( F \) is the constant force applied, \( t \) is the time duration the force acts, and \( \Delta p \) is the change in momentum of the object.

In the context of relativistic momentum, both the momentum and changes to it are influenced by the Lorentz factor. This means that \( \Delta p \) is not simply the mass times the change in velocity (as in classical mechanics) but instead incorporates the relavistic formula \( \Delta p = m \Delta(v \gamma) \).

This theorem is crucial when considering how long a force must be applied to bring an object to rest, especially at high speeds. The negative sign when stopping an object indicates the direction of the force is opposite to the motion. Understanding how to apply this theorem involves accurately determining \( \Delta p \) in the problem's context.

Newton's Second Law

Newton’s Second Law is one of the cornerstones of classical mechanics. It traditionally describes the relationship between the force applied to an object, its mass, and the acceleration that results. Stated simply, it is:

• \( F = ma \)

Where \( F \) is the force applied, \( m \) is mass, and \( a \) is acceleration.

However, when dealing with velocities approaching the speed of light, Newton's Second Law must be adapted into the relativistic framework. Here, instead of using \( F = ma \), we work with force as a function of momentum:

• \( F = \frac{dp}{dt} \)

In this form, \( dp \) is the change in relativistic momentum, making it vital to include the Lorentz factor. Thus, Newton’s laws are extended into relativistic territory by modifying how we look at momentum and force.

While classical physics works well at ordinary speeds, when dealing in relativistic momentum problems like the original exercise, it’s crucial to consider these relativistic adjustments to the timeless principles from Newton.